Solve {left(6-sqrt{left(6-xright)^2-4^2}right)}^2=x^2+4^2

$${ \left(6- \sqrt{ { \left(6-x \right) }^{ 2 } - { 4 }^{ 2 } } \right) }^{ 2 } = { x }^{ 2 } + { 4 }^{ 2 }$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-x\right)^{2}.

$$\left(6-\sqrt{36-12x+x^{2}-4^{2}}\right)^{2}=x^{2}+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\left(6-\sqrt{36-12x+x^{2}-16}\right)^{2}=x^{2}+4^{2}$$

Subtract 16 from 36 to get 20.

$$\left(6-\sqrt{20-12x+x^{2}}\right)^{2}=x^{2}+4^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\sqrt{20-12x+x^{2}}\right)^{2}.

$$36-12\sqrt{20-12x+x^{2}}+\left(\sqrt{20-12x+x^{2}}\right)^{2}=x^{2}+4^{2}$$

Calculate \sqrt{20-12x+x^{2}} to the power of 2 and get 20-12x+x^{2}.

$$36-12\sqrt{20-12x+x^{2}}+20-12x+x^{2}=x^{2}+4^{2}$$

Add 36 and 20 to get 56.

$$56-12\sqrt{20-12x+x^{2}}-12x+x^{2}=x^{2}+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$56-12\sqrt{20-12x+x^{2}}-12x+x^{2}=x^{2}+16$$

Subtract x^{2} from both sides.

$$56-12\sqrt{20-12x+x^{2}}-12x+x^{2}-x^{2}=16$$

Combine x^{2} and -x^{2} to get 0.

$$56-12\sqrt{20-12x+x^{2}}-12x=16$$

Subtract 56-12x from both sides of the equation.

$$-12\sqrt{20-12x+x^{2}}=16-\left(56-12x\right)$$

To find the opposite of 56-12x, find the opposite of each term.

$$-12\sqrt{20-12x+x^{2}}=16-56+12x$$

Subtract 56 from 16 to get -40.

$$-12\sqrt{20-12x+x^{2}}=-40+12x$$

Square both sides of the equation.

$$\left(-12\sqrt{20-12x+x^{2}}\right)^{2}=\left(-40+12x\right)^{2}$$

Expand \left(-12\sqrt{20-12x+x^{2}}\right)^{2}.

$$\left(-12\right)^{2}\left(\sqrt{20-12x+x^{2}}\right)^{2}=\left(-40+12x\right)^{2}$$

Calculate -12 to the power of 2 and get 144.

$$144\left(\sqrt{20-12x+x^{2}}\right)^{2}=\left(-40+12x\right)^{2}$$

Calculate \sqrt{20-12x+x^{2}} to the power of 2 and get 20-12x+x^{2}.

$$144\left(20-12x+x^{2}\right)=\left(-40+12x\right)^{2}$$

Use the distributive property to multiply 144 by 20-12x+x^{2}.

$$2880-1728x+144x^{2}=\left(-40+12x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-40+12x\right)^{2}.

$$2880-1728x+144x^{2}=1600-960x+144x^{2}$$

Subtract 1600 from both sides.

$$2880-1728x+144x^{2}-1600=-960x+144x^{2}$$

Subtract 1600 from 2880 to get 1280.

$$1280-1728x+144x^{2}=-960x+144x^{2}$$

Add 960x to both sides.

$$1280-1728x+144x^{2}+960x=144x^{2}$$

Combine -1728x and 960x to get -768x.

$$1280-768x+144x^{2}=144x^{2}$$

Subtract 144x^{2} from both sides.

$$1280-768x+144x^{2}-144x^{2}=0$$

Combine 144x^{2} and -144x^{2} to get 0.

$$1280-768x=0$$

Subtract 1280 from both sides. Anything subtracted from zero gives its negation.

$$-768x=-1280$$

Divide both sides by -768.

$$x=\frac{-1280}{-768}$$

Reduce the fraction \frac{-1280}{-768} to lowest terms by extracting and canceling out -256.

$$x=\frac{5}{3}$$

Substitute \frac{5}{3} for x in the equation \left(6-\sqrt{\left(6-x\right)^{2}-4^{2}}\right)^{2}=x^{2}+4^{2}.

$$\left(6-\sqrt{\left(6-\frac{5}{3}\right)^{2}-4^{2}}\right)^{2}=\left(\frac{5}{3}\right)^{2}+4^{2}$$

Simplify. The value x=\frac{5}{3} satisfies the equation.

$$\frac{169}{9}=\frac{169}{9}$$

Equation -12\sqrt{x^{2}-12x+20}=12x-40 has a unique solution.

$$x=\frac{5}{3}$$